Author Topic: EM Drive Developments Thread 1  (Read 794258 times)

Offline Rodal

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Re: EM Drive Developments
« Reply #2780 on: 10/29/2014 03:59 PM »
...

We should also model this

Quote from: Brady, March, White, et.al.
The null force testing indicated that there was an average null force of 9.6 micronewtons present in the as tested configuration. The presence of this null force was a result of the DC power current of 5.6 amps running in the power cable to the RF amplifier from the liquid metal contacts. This current causes the power cable to generate a magnetic field that interacts with the torsion pendulum magnetic damper system. The null test data is also shown in Fig. 20.

And the awful drift up and down of their "baseline".

Frobnicat, do you have any suggestion on how best to model this?  In which direction (A) torsional, B) swinging with largest rotary inertia, C) swinging with lowest rotary inertia)  is their "9.6 micronewton null force" ?
I don't think it could be torsional.  It looks (from their arrangement) like it is a B) swinging motion of the Faztek beams that gets measured as a torsional displacement because of the coupling between swinging and torsion.

If so, it should be entered into my model, rather as they do (in a sort of clumsy way) by subtracting it from their measured response (they assume linearity).
« Last Edit: 10/29/2014 06:03 PM by Rodal »

Offline Rodal

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Re: EM Drive Developments
« Reply #2781 on: 10/29/2014 06:26 PM »
Eagleworks inverted torsional pendulum response to exponentially decaying forcing functions (force in Newtons)

  Rodal 2014  :)


Piecewise[{{(80*10^(-6))*(1-Exp[-t/tau]),t<30},{0,t>= 30}}],tau=0,0.5,1,2,3

Piecewise[{{(80*10^(-6))*(1-Exp[-t/tau]),t<30},{(80*10^(-6))*(Exp[-(t-30)/tau]),t>= 30}}],tau=0,0.5,1,2,3

Piecewise[{{(80*10^(-6))*(1-Exp[-t/tau]),t<30},{(80*10^(-6))*(Exp[-(t-30)/tau]),t>= 30}}],tau=2

Here are the phase plots for

Piecewise[{{(80*10^(-6))*(1-Exp[-t/tau]),t<30},{(80*10^(-6))*(Exp[-(t-30)/tau]),t>= 30}}],tau=0.000001

and

Piecewise[{{(80*10^(-6))*(1-Exp[-t/tau]),t<30},{(80*10^(-6))*(Exp[-(t-30)/tau]),t>= 30}}],tau=2

It is evident that the numerical solution of this system of stiff coupled nonlinear equations is smooth

(Picture the solution of the differential equation as being a dot that follows these paths)
« Last Edit: 10/29/2014 06:38 PM by Rodal »

Offline aero

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Re: EM Drive Developments
« Reply #2782 on: 10/29/2014 06:29 PM »
I need a little help here. I'm trying to get an accurate estimate of the Brady cavity dimensions so that frobnicat will have the data he needs to evaluate dimensionally accurate equation formulations for Force. Using the attached left side photo, I have used my screen pixel ruler to extract dimensions as follows.

                   x -pixels   y -pixels   pixel dist
               w-b   3             246   246.018292
               w-s   14           138   138.7083271
                   L   221             0   221
------------------------------------------
top slope   -221   57   0.257918552
bot slope   213   54   -0.253521127
taper         0.511439679

Using the right side photo I have sketched what I think the camera sees. That is, it sees a chord of the big and small ends, and foreshortened length. To estimate the degree of foreshortening, I measured the cross section of the Faztek beams circled in red on the photos. The beam near the camera measures 41 pixels on a side, and the far beams measure 25 pixels wide. I measured both diagonals of the near beam end, and both widths of the far beams. Only one measurement differed by 1 pixel.

Assuming the center of the base is equidistant from the near and far beams, (Might need to adjust this slightly)I calculate the beam width of the Faztek beam supporting the cavity corresponding to the distance from camera to axial center of the cavity to be 33.125 pixels = 1.5 inches = 3.81 cm, or 0.115018868 cm/pixel. Using this conversion I calculate the chord lengths illustrated in the attached drawing to be
   w-big chord = 28.29674543 cm
w-small chord = 15.95407475 cm
and foreshortened length = 25.41916981 cm

But I need some help calculating the actual diameters and real length. It is not a huge factor but it is probably the largest error source remaining in the estimated cavity size.



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Offline Notsosureofit

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Re: EM Drive Developments
« Reply #2783 on: 10/29/2014 07:30 PM »
For what it's worth, I measure the FRP board at 0.060" and the copper cladding at 0.002". (the stuff I have here anyway)
« Last Edit: 10/29/2014 07:31 PM by Notsosureofit »

Offline aero

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Re: EM Drive Developments
« Reply #2784 on: 10/29/2014 07:43 PM »
For what it's worth, I measure the FRP board at 0.060" and the copper cladding at 0.002". (the stuff I have here anyway)
My measurements are taken at the inside bend of the copper. The FRC boards are outside that point, but the copper cladding is inside that point. IOW extend my length by 2 * 0.002" and shorten my diameters by the same amount. That is a further correction we should make.
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Offline Rodal

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Re: EM Drive Developments
« Reply #2785 on: 10/29/2014 08:37 PM »
I need a little help here. I'm trying to get an accurate estimate of the Brady cavity dimensions so that frobnicat will have the data he needs to evaluate dimensionally accurate equation formulations for Force. Using the attached left side photo, I have used my screen pixel ruler to extract dimensions as follows.

                   x -pixels   y -pixels   pixel dist
               w-b   3             246   246.018292
               w-s   14           138   138.7083271
                   L   221             0   221
------------------------------------------
top slope   -221   57   0.257918552
bot slope   213   54   -0.253521127
taper         0.511439679

Using the right side photo I have sketched what I think the camera sees. That is, it sees a chord of the big and small ends, and foreshortened length. To estimate the degree of foreshortening, I measured the cross section of the Faztek beams circled in red on the photos. The beam near the camera measures 41 pixels on a side, and the far beams measure 25 pixels wide. I measured both diagonals of the near beam end, and both widths of the far beams. Only one measurement differed by 1 pixel.

Assuming the center of the base is equidistant from the near and far beams, (Might need to adjust this slightly)I calculate the beam width of the Faztek beam supporting the cavity corresponding to the distance from camera to axial center of the cavity to be 33.125 pixels = 1.5 inches = 3.81 cm, or 0.115018868 cm/pixel. Using this conversion I calculate the chord lengths illustrated in the attached drawing to be
   w-big chord = 28.29674543 cm
w-small chord = 15.95407475 cm
and foreshortened length = 25.41916981 cm

But I need some help calculating the actual diameters and real length. It is not a huge factor but it is probably the largest error source remaining in the estimated cavity size.

This is not just shrinking Fornaro's dimensions but it is also changing the ratios:

Aero BigDiameter/SmallDiameter =  28.29674543 /  15.95407475 = 1.77

Fornaro BigDiameter/SmallDiameter = 39.7/24.4 = 1.63

Difference=9%

Aero CavityLength/SmallDiameter = 25.41916981 /  15.9540747 = 1.59

Fornaro CavityLength/SmallDiameter = 33.2 /24.4= 1.36

Difference=17%


Aero CavityLength/BigDiameter= 25.41916981 /  28.29674543 = 0.898

Fornaro CavityLength/BigDiameter= 33.2 /39.7 = 0.836

Difference= 7%

The most significant change is the ratio CavityLength/SmallDiameter
« Last Edit: 10/29/2014 08:54 PM by Rodal »

Offline frobnicat

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Re: EM Drive Developments
« Reply #2786 on: 10/29/2014 09:44 PM »
...

We should also model this

Quote from: Brady, March, White, et.al.
The null force testing indicated that there was an average null force of 9.6 micronewtons present in the as tested configuration. The presence of this null force was a result of the DC power current of 5.6 amps running in the power cable to the RF amplifier from the liquid metal contacts. This current causes the power cable to generate a magnetic field that interacts with the torsion pendulum magnetic damper system. The null test data is also shown in Fig. 20.

And the awful drift up and down of their "baseline".

Frobnicat, do you have any suggestion on how best to model this?  In which direction (A) torsional, B) swinging with largest rotary inertia, C) swinging with lowest rotary inertia)  is their "9.6 micronewton null force" ?
I don't think it could be torsional.  It looks (from their arrangement) like it is a B) swinging motion of the Faztek beams that gets measured as a torsional displacement because of the coupling between swinging and torsion.

If so, it should be entered into my model, rather as they do (in a sort of clumsy way) by subtracting it from their measured response (they assume linearity).

At the moment I do ask a lot of questions and don't answer a lot. Have to say I still don't have a good picture of the torsion pendulum geometry and where gizmos are relative to pivot (save the thruster).

Time to get a better understanding : from figure 1, left picture, we have a gantry that's static, along the vertical left "leg" of this gantry there are two flexure bearing with vertical axis, on the right picture we can only see the upper one, as two dark grey bloc and I guess the "spring foils" are hidden in the cylindrical space between them. Is that it ?

So essentially we have a vertical pivot on the left of the diamond shaped plate (second picture figure 17)  that's fixed on the right of the horizontal rotating arm, so that pivot axis is roughly centered in the middle of the faztek arm. This pivot has a linear restoring torque proportional to angular deviation from rest position. On the right of the upper part of vertical plate that links the arm to the flexure bearings there is a connecting box, this box will move with the arm. 

The static gantry is on three height adjustable (and vibration isolating ?) platforms so that the axis can be set a good vertical I guess. The oblique horizontal faztek linking the arm to the right leg of the gantry is seen on some pictures, not others; I guess it is used to fasten the rotating arm when mounting things and then removed for measures.

Figure 1 picture left : the liquid contacts system is the thing on top of the gantry, on the left, above the flexure bearings (aligned with axis of rotation ?). On figure 6 the upper white board is fixed to the gantry, the white board below (with cut angles) is the part that moves with the arm. It is somehow mechanically and electrically connected to the connecting box below.

We see the damping system on the third picture of figure 17 : it is situated at the back of the arm, below the amplifier. When the arm rotates, the "fin" enter and leaves the space between the strong magnets : figure 3 second picture the view is lateral to the arm, the plunging "fin" is fixed to back end of arm, it goes forward backward relative to the view. The permanent magnets are enclosed in a ferromagnetic U trying to close the circuit.

It's tempting to see the leaks of this magnetic circuit as a dipole with axis roughly aligned with arm. Hard to tell from the pictures (maybe with an added hour of eye straining ...) where is the cable that goes from the liquid contacts, above the axis, to the RF amplifier, back of the arm. Could be in the same plane, vertical plane defined by the arm. And magnetic field lines would be parallel to this plane (that needs to be checked, the U closing the magnetic circuit is not symmetric relative to that plane). That would give a cross product ILxB force directed orthogonal to that vertical plane above the arm.

Ok lets say, the arm is axis x positive front (thruster)  negative back (RF amplifier and magnets). The y axis is orthogonal, going to the right, the z axis is upward. And origin at the middle of the arm, at the axis of rotation. I would say the spurious DC force is along the y direction, applied somewhere between the wet contacts above the origin at x=y=0 z=+something and the amplifier at x=-something y=0 z=+not_much

If you could confirm my x y z link to your A B C :
A) torsional : rotation around z axis ?
B) swinging with largest rotary inertia : rotation around y axis ?
C) swinging with lowest rotary inertia) : rotation around x axis ?

The B and C modes would be very stiff (similar stiffness)
The A and B modes would have similar moment of inertia
The flexure bearings would introduce some level of coupling between the 3 angles + 3 displacements, is that where you get the nonlinearities ?

From this line of reasonning the DC spurious force would have a direct torque around the z (mode A) and x (mode C) axis. Just the opposite of what you said :)

Only... isn't a DC current supposed to go and come back ? A twisted pair would in principle suffice to neutralize any net imbalance. Only when the cables separate we have a loop with coupling to magnetic field. So, where the cables separate (at the wet contacts box? inside the amplifier?) and at what angle in what plane ? Short of those answers, best guess is to take into account only what goes into A mode and discard the chaotic aspects of A B C coupling.

So, in the end, for the principal dynamic activity of the balance around A, at 99% we have a simple (under)damped harmonic oscillator no ? What is the force vs speed function of a magnetic damper ?




Offline frobnicat

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Re: EM Drive Developments
« Reply #2787 on: 10/29/2014 10:01 PM »

Well, there is a lot to answer there.  But what do you think of just modeling the impulse as a trapezoid ?

That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?

then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2

From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.

I understand you see a piecewise linear function as the default way to introduce more (not too much) parameters to fit the target data. It's ok for me.

From my "more complicated picture" that will take many hours to just utter  (a chance that you are patient !) I still suggest rectangle plus exp "charge/discharge". If piecewise linear it would rather look like :

     ______
    /      |
____|      \_______


I'm a big fan of chaotic swing too, but right now I'm on an agenda with lower Lyapunov exponent.

Offline Ron Stahl

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Re: EM Drive Developments
« Reply #2788 on: 10/29/2014 10:02 PM »
Only... isn't a DC current supposed to go and come back ? A twisted pair would in principle suffice to neutralize any net imbalance. Only when the cables separate we have a loop with coupling to magnetic field. So, where the cables separate (at the wet contacts box? inside the amplifier?) and at what angle in what plane ?
Since they claim to know they have b field coupling, one presumes they waved some Mu metal between or some such.  It wouldn't be the first time a twisted pair produced a noticeable force and these are very small forces.
« Last Edit: 10/29/2014 10:03 PM by Ron Stahl »

Offline Rodal

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Re: EM Drive Developments
« Reply #2789 on: 10/29/2014 10:59 PM »
...

We should also model this

Quote from: Brady, March, White, et.al.
The null force testing indicated that there was an average null force of 9.6 micronewtons present in the as tested configuration. The presence of this null force was a result of the DC power current of 5.6 amps running in the power cable to the RF amplifier from the liquid metal contacts. This current causes the power cable to generate a magnetic field that interacts with the torsion pendulum magnetic damper system. The null test data is also shown in Fig. 20.

And the awful drift up and down of their "baseline".

Frobnicat, do you have any suggestion on how best to model this?  In which direction (A) torsional, B) swinging with largest rotary inertia, C) swinging with lowest rotary inertia)  is their "9.6 micronewton null force" ?
I don't think it could be torsional.  It looks (from their arrangement) like it is a B) swinging motion of the Faztek beams that gets measured as a torsional displacement because of the coupling between swinging and torsion.

If so, it should be entered into my model, rather as they do (in a sort of clumsy way) by subtracting it from their measured response (they assume linearity).

At the moment I do ask a lot of questions and don't answer a lot. Have to say I still don't have a good picture of the torsion pendulum geometry and where gizmos are relative to pivot (save the thruster).

Time to get a better understanding : from figure 1, left picture, we have a gantry that's static, along the vertical left "leg" of this gantry there are two flexure bearing with vertical axis, on the right picture we can only see the upper one, as two dark grey bloc and I guess the "spring foils" are hidden in the cylindrical space between them. Is that it ?

So essentially we have a vertical pivot on the left of the diamond shaped plate (second picture figure 17)  that's fixed on the right of the horizontal rotating arm, so that pivot axis is roughly centered in the middle of the faztek arm. This pivot has a linear restoring torque proportional to angular deviation from rest position. On the right of the upper part of vertical plate that links the arm to the flexure bearings there is a connecting box, this box will move with the arm. 

The static gantry is on three height adjustable (and vibration isolating ?) platforms so that the axis can be set a good vertical I guess. The oblique horizontal faztek linking the arm to the right leg of the gantry is seen on some pictures, not others; I guess it is used to fasten the rotating arm when mounting things and then removed for measures.

Figure 1 picture left : the liquid contacts system is the thing on top of the gantry, on the left, above the flexure bearings (aligned with axis of rotation ?). On figure 6 the upper white board is fixed to the gantry, the white board below (with cut angles) is the part that moves with the arm. It is somehow mechanically and electrically connected to the connecting box below.

We see the damping system on the third picture of figure 17 : it is situated at the back of the arm, below the amplifier. When the arm rotates, the "fin" enter and leaves the space between the strong magnets : figure 3 second picture the view is lateral to the arm, the plunging "fin" is fixed to back end of arm, it goes forward backward relative to the view. The permanent magnets are enclosed in a ferromagnetic U trying to close the circuit.

It's tempting to see the leaks of this magnetic circuit as a dipole with axis roughly aligned with arm. Hard to tell from the pictures (maybe with an added hour of eye straining ...) where is the cable that goes from the liquid contacts, above the axis, to the RF amplifier, back of the arm. Could be in the same plane, vertical plane defined by the arm. And magnetic field lines would be parallel to this plane (that needs to be checked, the U closing the magnetic circuit is not symmetric relative to that plane). That would give a cross product ILxB force directed orthogonal to that vertical plane above the arm.

Ok lets say, the arm is axis x positive front (thruster)  negative back (RF amplifier and magnets). The y axis is orthogonal, going to the right, the z axis is upward. And origin at the middle of the arm, at the axis of rotation. I would say the spurious DC force is along the y direction, applied somewhere between the wet contacts above the origin at x=y=0 z=+something and the amplifier at x=-something y=0 z=+not_much

If you could confirm my x y z link to your A B C :
A) torsional : rotation around z axis ?
B) swinging with largest rotary inertia : rotation around y axis ?
C) swinging with lowest rotary inertia) : rotation around x axis ?

The B and C modes would be very stiff (similar stiffness)
The A and B modes would have similar moment of inertia
The flexure bearings would introduce some level of coupling between the 3 angles + 3 displacements, is that where you get the nonlinearities ?

From this line of reasonning the DC spurious force would have a direct torque around the z (mode A) and x (mode C) axis. Just the opposite of what you said :)

Only... isn't a DC current supposed to go and come back ? A twisted pair would in principle suffice to neutralize any net imbalance. Only when the cables separate we have a loop with coupling to magnetic field. So, where the cables separate (at the wet contacts box? inside the amplifier?) and at what angle in what plane ? Short of those answers, best guess is to take into account only what goes into A mode and discard the chaotic aspects of A B C coupling.

So, in the end, for the principal dynamic activity of the balance around A, at 99% we have a simple (under)damped harmonic oscillator no ? What is the force vs speed function of a magnetic damper ?

the arm is axis x positive front (thruster)  negative back (RF amplifier and magnets).
The y axis is orthogonal, going to the right,
the z axis is upward.

alpha=rotation around x axis
beta=rotation around y axis
gamma=rotation around z axis


OK let me answer one important question at the outset: where does the coupling come from.

The coupling comes from this nasty fact:

if you have a force applied at the origin along the x axis, it will be produce a swinging beta rotation around the y axis

if you have a force applied at the origin along the y axis, it will be produce a swinging alpha rotation around the x axis


if you have a force applied at the end of the x arm, oriented along the y axis, it will produce a gamma rotation around the z axis, but also (because of the above facts) one has nonlinear coupling:

alphaDot =  d alpha /dt
alphaDotDot = d2  alpha /dt2

In the equations of motion for the gamma rotation around the z axis one also gets contributions from a number of terms, the most important ones being the following rates:  alphaDot * betaDot  and another contribution from beta * alphaDotDot

I obtained the 3-dimensional, nonlinearly coupled equations of motion by solving the Lagrangian.

Please notice that while the department of Aeronautics and Astronautics at MIT has an inverted pendulum designed at MIT to eliminate this coupling (only linear x and y motions are allowed for the thruster), NASA Eagleworks neglected to eliminate this coupling. 

The department of Aeronautics and Astronautics at MIT has been a leader in nonlinear dynamics for the last century (starting with the problems of flutter and divergence and self-excited oscillations).
« Last Edit: 10/29/2014 11:16 PM by Rodal »

Offline Rodal

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Re: EM Drive Developments
« Reply #2790 on: 10/29/2014 11:29 PM »

Well, there is a lot to answer there.  But what do you think of just modeling the impulse as a trapezoid ?

That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?

then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2

From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.

I understand you see a piecewise linear function as the default way to introduce more (not too much) parameters to fit the target data. It's ok for me.

From my "more complicated picture" that will take many hours to just utter  (a chance that you are patient !) I still suggest rectangle plus exp "charge/discharge". If piecewise linear it would rather look like :

     ______
    /      |
____|      \_______


I'm a big fan of chaotic swing too, but right now I'm on an agenda with lower Lyapunov exponent.

f=0   @ t=0
f=f1  @ t=t1
f=f2  @ t=t2
f=f2  @ t=t3
f=f3  @ t=t4
f=0   @ t=t5

Please give me f1, f2,f3 and t1,t2,t3,t4,t5 you would like

f1 = f2 - f3  therefore f3 = f2 - f1  ?

t1 = t4 - t3 therefore t4 = t1 + t3  ?

t2 - t1 = t5 - t4 therefore t5 = t2 + t3  ?

therefore,

need f1, f2,  and t1,t2,t3


« Last Edit: 10/30/2014 01:06 AM by Rodal »

Offline Rodal

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Re: EM Drive Developments
« Reply #2791 on: 10/30/2014 02:34 AM »

Offline Rodal

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Re: EM Drive Developments
« Reply #2792 on: 10/30/2014 02:34 AM »
....
From this line of reasonning the DC spurious force would have a direct torque around the z (mode A) and x (mode C) axis. Just the opposite of what you said :)
....

Based on my memory of their setup, I am not following your line of reasoning here.  Let me go back to Brady's report and check whether I remember correctly their setup.

Figure 3 shows the magnetic damper

One can see a porthole that is on a line perpendicular to the beam attached to the magnetic damper

The portholes are on the sides of the chamber

Therefore the magnetic damper is connected to the beam that runs longitudinally along the length of the chamber.  Therefore the magnetic damper is connected to a beam that runs along the x axis.

If that beam running along the x axis goes through the center of rotation, the force oriented along direction x cannot produce a torque along the z axis, because the direction of the force will go through the center of torsional rotation.

A force directed along the x axis produces a swinging rotation "beta" around the y axis,  swinging with largest rotary inertia.


                                                                                         <-------------  X direction

                     

Y axis perpendicular to this page     


                                                                                             

           <--
       /           \                                                                 
beta rotation l                       
« Last Edit: 10/30/2014 02:39 AM by Rodal »

Offline aero

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Re: EM Drive Developments
« Reply #2793 on: 10/30/2014 03:14 AM »
I'm getting really tired of looking at these pictures so I'm going to join John after I post what I've found today.

See http://forum.nasaspaceflight.com/index.php?topic=29276.msg1279371#msg1279371 for background and photos.

Pinhole projection formula         x/f = X/S   
where            
x1 = object size on photo 1         25   px
x2 = object size on photo 2         41   px
f = focal length            
X = object real size               3.81   cm
d1 = distance from camera to object, photo 1            
S = distance between 1 and 2         31.12641998   cm
which I assumed as 1.1 times the w-big chord.
then            
d1=S*x2/(x2-x1)            
   d1=   79.76145119   cm   so the distance to the center of the base plate, dc, is d1 - 0.6* w-big chord
dc = 62.78340393 cm
            
With that, using alpha = 2* arctan(w-big chord/2*dc) you can calculate the angle subtended by the w-big chord then it is just a matter of using right triangles to calculate the real radius, big and small, then scale the length proportionately. Note that this will make a very noticeable difference in cavity size.

Errors that I know about: When I calculated the scale factor I assumed the two reference beams were equidistant from the center of the base. Closer inspection leads me to think that the beam in front is aligned with the front of edge of the base, but the beam in back is separated from the edge of the base by about 0.1 diameters of the base. I have used that observation here, but it needs to be looked at by younger eyes then fitted into the estimate of the scale factor made in my previous post. That will adjust all dimensions of the cavity. After that, re-do this part of course.

Maybe I'll feel like doing more after I get some feed back regarding beam location and separation estimated in units of big end diameters. Or cm, if you like.
Retired, working interesting problems

Offline ThinkerX

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Re: EM Drive Developments
« Reply #2794 on: 10/30/2014 04:04 AM »
I am starting to wonder if the 'EM Drive' is not so much 'generating thrust' but rather trying to 'hitch a ride' on the expansive force of the universe, be it 'Dark Energy' or the 'Cosmological Constant' or something else. 

The tiny bit of info I could grasp about Dark Energy is astonishing: a force large enough to overcome the gravitational attraction between galaxies (!) forcing them apart at a continuous acceleration on the order of 1 km/sec per kilo parsec (!)  A force that grows over time rather than diminishes?   Regardless of the exact mechanism, the observed astronomical effects are real on the macroscopic scale, which to me implies they must be real at the local level as well, if much diminished for whatever reason. 

From the tiny bit of the papers I could actually understand, it seems like the theorist are highly uncomfortable with the measures required to fit Dark Energy, in any form, into the scheme of physics.  Multiple tepid explanations. 

One oddity:  Despite its name, the 'Cosmological Constant' invoked to help explain Dark Energy is a variable, not a fixed number, PLUS one of the major EM Drives is that the frequency the various experimental groups are trying to hit is also a variable.  Wonder if there could be a direct or indirect connection there: the experimenters hit the right frequency in a container with the right dimensions, and Dark Energy comes calling.  Or did I just reinvent the wheel here?

Almost like Dark Energy doesn't play well with other universal forces, which tells me about the only way to verify this effect would be in space - preferably way out in space, away from other influences.

Bizarre hypothetical or question:  if you could somehow generate Dark Energy inside a sealed container, would the laws of thermodynamics, especially conservation of motion, still apply?  Would they still apply if Dark Energy was interacting with another force inside the sealed container?  Or did I just reinvent the wheel again?

Ok...time for the rest of you to post the cute videos demonstrating how deep of a hole I dug for myself this time...
« Last Edit: 10/30/2014 04:40 AM by ThinkerX »

Offline frobnicat

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Re: EM Drive Developments
« Reply #2795 on: 10/30/2014 08:29 AM »

Well, there is a lot to answer there.  But what do you think of just modeling the impulse as a trapezoid ?

That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?

then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2

From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.

I understand you see a piecewise linear function as the default way to introduce more (not too much) parameters to fit the target data. It's ok for me.

From my "more complicated picture" that will take many hours to just utter  (a chance that you are patient !) I still suggest rectangle plus exp "charge/discharge". If piecewise linear it would rather look like :

     ______
    /      |
____|      \_______


I'm a big fan of chaotic swing too, but right now I'm on an agenda with lower Lyapunov exponent.

f=0   @ t=0
f=f1  @ t=t1
f=f2  @ t=t2
f=f2  @ t=t3
f=f3  @ t=t4
f=0   @ t=t5

Please give me f1, f2,f3 and t1,t2,t3,t4,t5 you would like

f1 = f2 - f3  therefore f3 = f2 - f1  ?

t1 = t4 - t3 therefore t4 = t1 + t3  ?

t2 - t1 = t5 - t4 therefore t5 = t2 + t3  ?

therefore,

need f1, f2,  and t1,t2,t3

Not much time, a mountain is waiting for my feet :
f=0   @ t=0-
f=80(1-ssr)  @ t=0+  (instantaneous rise)
f=80  @ t=tau
f=80  @ t=30-
f=80ssr  @ t=30+ (instantaneous fall)
f=0   @ t=30+tau

only two free running parameters : tau for the slow component and ssr = smooth/sharp ratio
let's try  ssr = .4  .2  .1
together with
tau = 0.25  0.5  1   2   4

more later about x y z
« Last Edit: 10/30/2014 08:32 AM by frobnicat »

Offline JohnFornaro

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Re: EM Drive Developments
« Reply #2796 on: 10/30/2014 11:13 AM »
Ok...time for the rest of you to post the cute videos demonstrating how deep of a hole I dug for myself this time...

We aim to please:

Sometimes I just flat out don't get it.

Offline Rodal

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Re: EM Drive Developments
« Reply #2797 on: 10/30/2014 03:45 PM »

Well, there is a lot to answer there.  But what do you think of just modeling the impulse as a trapezoid ?

That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?

then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2

From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.

I understand you see a piecewise linear function as the default way to introduce more (not too much) parameters to fit the target data. It's ok for me.

From my "more complicated picture" that will take many hours to just utter  (a chance that you are patient !) I still suggest rectangle plus exp "charge/discharge". If piecewise linear it would rather look like :

     ______
    /      |
____|      \_______


I'm a big fan of chaotic swing too, but right now I'm on an agenda with lower Lyapunov exponent.

f=0   @ t=0
f=f1  @ t=t1
f=f2  @ t=t2
f=f2  @ t=t3
f=f3  @ t=t4
f=0   @ t=t5

Please give me f1, f2,f3 and t1,t2,t3,t4,t5 you would like

f1 = f2 - f3  therefore f3 = f2 - f1  ?

t1 = t4 - t3 therefore t4 = t1 + t3  ?

t2 - t1 = t5 - t4 therefore t5 = t2 + t3  ?

therefore,

need f1, f2,  and t1,t2,t3

Not much time, a mountain is waiting for my feet :
f=0   @ t=0-
f=80(1-ssr)  @ t=0+  (instantaneous rise)
f=80  @ t=tau
f=80  @ t=30-
f=80ssr  @ t=30+ (instantaneous fall)
f=0   @ t=30+tau

only two free running parameters : tau for the slow component and ssr = smooth/sharp ratio
let's try  ssr = .4  .2  .1
together with
tau = 0.25  0.5  1   2   4

more later about x y z

Results for

Piecewise[{
{(80*10^(-6))*(1 - ssr (1 - t/tau)), t <= tau}, {(80*10^(-6)), t <  30},
{(80*10^(-6))*(ssr/tau)*(30 + tau - t), t <= 30 + tau}
                }]

We see the effect of tau and ssr modifying mainly the dynamic magnification factor, as expected.


Varying tau at constant ssr

ssr=0.4 tau = 0, 1,2,3,4,5

ssr=0.4 tau = 4

Note: there are diminishing returns on this effect for tau >3



Varying ssr at constant tau

ssr=0, 0.25, 0.50, 0.75, 1.00 tau = 4

ssr=0.5 tau = 4

Note: ssr close to 1 effectively "kills" the dynamic magnification factor (the "first ring" in Fornaro vernacular) without much affecting the second ring



« Last Edit: 10/30/2014 09:42 PM by Rodal »

Offline Rodal

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Re: EM Drive Developments
« Reply #2798 on: 10/30/2014 06:29 PM »
Ok...time for the rest of you to post the cute videos demonstrating how deep of a hole I dug for myself this time...

We aim to please:

Can you see a sextupole perturbation in the resonant excitation of the EM Drive ?

Offline Rodal

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Re: EM Drive Developments
« Reply #2799 on: 10/30/2014 08:21 PM »

Well, there is a lot to answer there.  But what do you think of just modeling the impulse as a trapezoid ?

That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?

then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2

From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.

I understand you see a piecewise linear function as the default way to introduce more (not too much) parameters to fit the target data. It's ok for me.

From my "more complicated picture" that will take many hours to just utter  (a chance that you are patient !) I still suggest rectangle plus exp "charge/discharge". If piecewise linear it would rather look like :

     ______
    /      |
____|      \_______


I'm a big fan of chaotic swing too, but right now I'm on an agenda with lower Lyapunov exponent.

f=0   @ t=0
f=f1  @ t=t1
f=f2  @ t=t2
f=f2  @ t=t3
f=f3  @ t=t4
f=0   @ t=t5

Please give me f1, f2,f3 and t1,t2,t3,t4,t5 you would like

f1 = f2 - f3  therefore f3 = f2 - f1  ?

t1 = t4 - t3 therefore t4 = t1 + t3  ?

t2 - t1 = t5 - t4 therefore t5 = t2 + t3  ?

therefore,

need f1, f2,  and t1,t2,t3

Not much time, a mountain is waiting for my feet :
f=0   @ t=0-
f=80(1-ssr)  @ t=0+  (instantaneous rise)
f=80  @ t=tau
f=80  @ t=30-
f=80ssr  @ t=30+ (instantaneous fall)
f=0   @ t=30+tau

only two free running parameters : tau for the slow component and ssr = smooth/sharp ratio
let's try  ssr = .4  .2  .1
together with
tau = 0.25  0.5  1   2   4

more later about x y z

Results for

Piecewise[{{(80*10^(-6))*(1 - ssr (1 - t/tau)),
   t <= tau}, {(80*10^(-6)),
   t <  30}, {(80*10^(-6))*(ssr/tau)*(30 + tau - t), t <= 30 + tau}}]

We see the effect of tau and ssr modifying mainly the dynamic magnification factor, as expected.


Varying tau at constant ssr

ssr=0.4 tau = 0, 1,2,3,4,5

ssr=0.4 tau = 4

Note: there are diminishing returns on this effect for tau >3



Varying ssr at constant tau

ssr=0, 0.25, 0.50, 0.75, 1.00 tau = 4

ssr=0.5 tau = 4

Note: ssr close to 1 effectively "kills" the dynamic magnification factor (the "first ring" in Fornaro vernacular) without much affecting the second ring



Here is the decaying exponential study:

Piecewise[{
{(80*10^(-6))*(1-ssr)+(80*10^(-6))*ssr*(1-Exp[-t/tau])/(1-Exp[-30/tau]),t<30},
{(80*10^(-6))*ssr*(Exp[-(t-30)/tau]),t>= 30}
                }]

Varying ssr at constant tau

ssr=0, 0.25, 0.50, 0.75, 1.00 tau = 4

ssr=0.5 tau = 4

Note: ssr close to 1

A) The piecewise linear forcing function effectively "kills" the dynamic magnification factor (the "first ring") without much affecting the second ring

B) The exponential decay forcing function (controlling only the initial force through ssr) kills the dynamic magnification factor as well as affecting the other rings




Varying tau at constant ssr

Piecewise[{
{(80*10^(-6))*(1-ssr)+(80*10^(-6))*ssr*(1-Exp[-t/tau])/(1-Exp[-30/tau]),t<30},
{(80*10^(-6))*ssr*(Exp[-(t-30)/tau]),t>= 30}
               }]

ssr=0.4 tau = 0, 1,2,3,4,5

ssr=0.4 tau = 4

« Last Edit: 10/30/2014 09:38 PM by Rodal »

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